Insertion Sort
Which are true of Insertion Sort (traditional implementation, without optimizations)?
Multiple answers:You can select more than one option.
Please, include the explanation with the answer.
A) It uses Θ(n^2) comparisons in the worst case
B) It uses Θ(n^2) comparisons in the average case
C) It uses Θ(n^2) comparisons in the best case
D) It uses Θ(n^2) movements of elements in the worst case
E) It uses Θ(n^2) movements of elements in the average case
F) It uses Θ(n^2) movements of elements in the best case
Two correct options should be: A and C
Reason: Counting comparisons or swaps(movement of elements) yields similar results. Each time through the inner for loop yields both a comparison and a swap, except the last (i.e., the comparison that fails the inner for loop's test), which has no swap. Thus, the number of swaps for the entire sort operation is n−1 less than the number of comparisons. This is 0 in the best case, and O(n^2) in the average and worst cases.
For reference Insertion sort algorithm goes like:
INSERTION-SORT(A) for i = 1 to n //outer loop run n times in best case key = A [i] j = i – 1 while j > = 0 and A[j] > key //inner loop for comparison & swapping A[j+1] = A[j] j = j – 1 End while A[j+1] = key End for
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Insertion Sort Which are true of Insertion Sort (traditional implementation, without optimizations)? Multiple answers:You can select...
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